Unformatted text preview: v (t), the answer
is v (2) = 3m/sec.
2. x = 2t2 + 3 and y = t4 . Find dy
d2 y
and 2 when t = −1.
dx
dx
dy
=
dx dy
dt
dx
dt = 4t3
= t2
4t dy
Therefore,
t=−1 = 1.
dx
Next,
d2 y
d dy
=
( )=
dx2
dx dx d dy
()
dt dx
dx
dt = 2t
1
=.
4t
2 d2 y
1
Therefore, 2 t=−1 = .
dx
2
3. Find the equation of the line normal to the curve x2 y 2 = 9 at the point (−1, 3).
By implicit diﬀerentiation and the Product rule,
dy
=0
dx
dy
x2 2 y
= −2xy 2
dx
dy
−2xy 2
=2
.
dx
x 2y 2xy 2 + x2 2y dy
Therefore, the slope of the line tangent to the curve at (−1, 3) is
(−1,3) = 3. We are,
dx
however, looking for the equation of the line normal to the curve at this point, so the
1 2 1
slope of this line will be − , and the normal line equation is
3
1
y − 3 = − (x + 1).
3...
View
Full
Document
This note was uploaded on 02/10/2014 for the course MATH 235 taught by Professor Lucas during the Fall '08 term at James Madison University.
 Fall '08
 LUCAS
 Calculus

Click to edit the document details